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kristall2001
@kristall2001
August 2022
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Дифференцируема ли функция y=|x^2-1|/|1-x| в области ее определения?
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flsh
Y = |x² - 1|/|1 - x|
D(y): x ≠ 1
y = |(x² - 1)/(1 - x)| = |-(x + 1)| = |x + 1|
(x = -1) ∈ D(y), но в ней функция недифференцируема, т. к. y’(x→-1⁻) ≠ y’(x→-1⁺)
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Answers & Comments
D(y): x ≠ 1
y = |(x² - 1)/(1 - x)| = |-(x + 1)| = |x + 1|
(x = -1) ∈ D(y), но в ней функция недифференцируема, т. к. y’(x→-1⁻) ≠ y’(x→-1⁺)